A bullet of mass 10 ~g is fired horizontally | vedclass

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(B) The work-energy theorem states that the work done by all forces equals the change in kinetic energy.$W_{total} = \Delta K = K_f - K_i$Here,the tot

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$3755$

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(B) The work-energy theorem states that the work done by all forces equals the change in kinetic energy.$W_{total} = \Delta K = K_f - K_i$Here,the total work done is the sum of work done by gravity $(W_g)$ and work done against air resistance $(W_{air})$.$W_g + W_{air} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$Given:Mass $m = 10 ~g = 0.01 ~kg$Initial velocity $u = 1000 ~ms^{-1}$Final velocity $v = 500 ~ms^{-1}$Height $h = 50 ~m$Gravity $g = 10 ~ms^{-2}$Work done by gravity $W_g = mgh = 0.01 	imes 10 	imes 50 = 5 ~J$Change in kinetic energy $\Delta K = \frac{1}{2} 	imes 0.01 	imes (500^2 - 1000^2)$$\Delta K = 0.005 	imes (250000 - 1000000) = 0.005 	imes (-750000) = -3750 ~J$Now,$W_g + W_{air} = \Delta K$$5 + W_{air} = -3750$$W_{air} = -3750 - 5 = -3755 ~J$The work done against air resistance is the magnitude of the work done by air resistance,which is $3755 ~J$.

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